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Question Number 86111 by john santu last updated on 27/Mar/20

$$\sin \left(\frac{3\pi }{2}\cos x\right)=-\frac{1}{2}$$

Commented by jagoll last updated on 27/Mar/20

$$\iff \sin \left(\frac{3\pi }{2}\cos \mathrm{x}\right)=\sin (-\frac{\pi }{6})$$$$\frac{3\pi }{2}\cos \mathrm{x}=-\frac{\pi }{6}+2\mathrm{k}\pi$$$$\cos \mathrm{x}=\frac{2}{3\pi }\{-\frac{\pi }{6}+2\mathrm{k}\pi \}$$$$\cos \mathrm{x}=-\frac{1}{9}+\frac{4\mathrm{k}}{3}$$$$\mathrm{x}={\cos }^{-1}\left(\frac{12\mathrm{k}-1}{9}\right)+2\mathrm{n}\pi$$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

$$sin\left(\frac{3\pi }{2}cosx\right)=sin(\pi +\frac{\pi }{6})$$$$\frac{3\pi }{2}cosx=\frac{7\pi }{6}$$$$cosx=\frac{7}{6}\times \frac{2}{3}=\frac{7}{9}=cos\alpha$$$$x=2n\pi \pm \alpha [\alpha ={cos}^{-1}\left(\frac{7}{9}\right)]$$$$★sin\left(\frac{3\pi }{2}cosx\right)=-\frac{1}{2}=sin\left(\frac{-\pi }{6}\right)$$$$\frac{3\pi }{2}cosx=\frac{-\pi }{6}cosx=\frac{-1}{9}=cos\beta$$$$x=2n\pi \pm \beta [\beta ={cos}^{-1}\left(\frac{-1}{9}\right)]$$$$$$

Answered by mr W last updated on 27/Mar/20

$$\sin \left(\frac{3\pi }{2}\cos x\right)=-\frac{1}{2}$$$$\Rightarrow \frac{3\pi }{2}\cos x=n\pi -{(-1)}^n\frac{\pi }{6}$$$$\Rightarrow \cos x=\frac{2}{3}[n-{(-1)}^n\frac{1}{6}]$$$$\Rightarrow -1⩽\frac{2}{3}[n-{(-1)}^n\frac{1}{6}]⩽1$$$$\Rightarrow n=-1,0,1$$$$\Rightarrow \cos x=\frac{2}{3}[n-{(-1)}^n\frac{1}{6}]=-\frac{5}{9},-\frac{1}{9},\frac{7}{9}$$$$\Rightarrow x=(2k+1)\pi \pm {\cos }^{-1}\frac{5}{9}$$$$\Rightarrow x=(2k+1)\pi \pm {\cos }^{-1}\frac{1}{9}$$$$\Rightarrow x=2k\pi \pm {\cos }^{-1}\frac{7}{9}$$

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